Electrons with a kinetic energy of 6.023 time | vedclass

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Question

(A) The kinetic energy $(KE)$ of $1 \ mol$ of electrons is $6.023 	imes 10^4 \ J$. Since $1 \ mol = 6.023 	imes 10^{23} \ 	ext{atoms}$,the $KE$ of

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$2.3125 	imes 10^{-19} \ J$

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(A) The kinetic energy $(KE)$ of $1 \ mol$ of electrons is $6.023 	imes 10^4 \ J$. Since $1 \ mol = 6.023 	imes 10^{23} \ 	ext{atoms}$,the $KE$ of $1 \ 	ext{electron}$ is: $KE = \frac{6.023 	imes 10^4 \ J}{6.023 	imes 10^{23}} = 1.0 	imes 10^{-19} \ J$. The energy of the incident photon $(E)$ is given by $E = \frac{hc}{\lambda}$: $E = \frac{6.626 	imes 10^{-34} \ J \cdot s 	imes 3 	imes 10^8 \ m/s}{600 	imes 10^{-9} \ m} = 3.313 	imes 10^{-19} \ J$. The threshold energy $(Phi)$ is the difference between the incident photon energy and the kinetic energy: $\Phi = E - KE = 3.313 	imes 10^{-19} \ J - 1.0 	imes 10^{-19} \ J = 2.313 	imes 10^{-19} \ J$.

Energy level	$K$ $(n=1)$	$L$ $(n=2)$	$M$ $(n=3)$	$N$ $(n=4...n=\infty)$
Energy	$-864 \ a.u.$	$-216 \ a.u.$	$-96 \ a.u.$	$0 \ a.u.$

Electrons with a kinetic energy of $6.023 \times 10^4 \ J/mol$ are evolved from the surface of a metal when it is exposed to radiation of wavelength $600 \ nm$. The minimum amount of energy required to remove an electron from the metal atom is

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